2
$\begingroup$

If I had a function $f(x,y)$ where $x=x(s,t)$ and $y=y(s,t)$ then

$$ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}. $$

I have been reading this and have spotted in case 2 that

$$ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}. $$

$f$ has no explicit $t$ dependence so how can this be correct? How can I have a chain rule for partial derivatives? A chain rule implies that there is no explicit dependence on the variable that we are differentiating with respect to, and I thought partial derivatives only deal with explicit dependencies and not implicit? Thanks

$\endgroup$
  • 2
    $\begingroup$ Notice that in your first equation two different functions are being called by the same name $f$. We should really define $\hat f(s,t) = f(x(s,t),y(s,t))$. Then the chain rule states that $\frac{\partial \hat f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}$. To be even more clear we could write $D_2 \hat f(s,t) = D_1 f(x(s,t),y(s,t)) D_2 x(s,t) + D_2 f(x(s,t),y(s,t)) D_2 y(s,t)$. For some reason calling $f$ and $\hat f$ by the same name is a common abuse of notation, but it causes a lot of confusion. $\endgroup$ – littleO Jul 11 '17 at 13:51
3
$\begingroup$

In case $2$, it's not $\frac{\partial f}{\partial t}$, it's $\frac{\partial z}{\partial t}$, and $z$ is not independent of $t$.

$\endgroup$
  • $\begingroup$ $f = z$ so the derivatives are identical? $\endgroup$ – Matt0410 Jul 11 '17 at 13:48
  • 2
    $\begingroup$ @Matt0410 No, $f = z$ is false. They are different functions. $z(s,t) = f(x(s,t),y(s,t))$. $\endgroup$ – littleO Jul 11 '17 at 13:54
  • 1
    $\begingroup$ @Matt0410 No, $f$ is a function of $x$ and $y$, while $z$ is a function of $s$ and $t$. For the $\partial$-notation, they are two different things. $\endgroup$ – 5xum Jul 11 '17 at 13:55
0
$\begingroup$

In a strict mathematical sense, you are correct. BUT: Engineers, physicists and other people who depend highly on calculus depend to simplify things, or drop parts of the standard notation.

In this case they mean $$\frac{\partial f(x,y)}{\partial t}(s,t) = \frac{\partial f}{\partial x}(x(s,t),y(s,t)) \frac{\partial x}{\partial t}(s,t) + \frac{\partial f}{\partial y}(x(s,t),y(s,t)) \frac{\partial y}{\partial t}(s,t) $$

As you can see, this is quite a lot to write (I hope, I did no typo's) Mathematicians tend to drop the $(s,t)$ and then do not mean that the evaluations of the functions are the same, but the functions are the same. Physicists (and other people mentioned obove) tend to even drop the dependency of $f$ and by that kind of overwrite the meaning of the old f by the composition with the new f....

$\endgroup$
-1
$\begingroup$

An ordinary derivative--such as that of $f$ with respect to $t$, in your example is only valid if all the dependencies of $f$, all the dependencies of those dependencies, all the dependencies of those dependencies, and so on, can be shown to have a single common dependency $t$, explicit or otherwise. In all other cases, only a partial derivative applies, since $f$ then only depends partially on $t$.

That is why a partial derivative is used in Case 2; $t$ is not a common dependency of the system $(f,x,y,s,t)$, as $s$ is in no way dependent upon $t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.